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# Suppose $f$ is a continuous positive decreasing function for $x \ge 1$ and $a_n = f(n).$ By drawing a picture, rank the following three quantities in increasing order. $\int^6_1 f(x) dx \displaystyle \sum_{i = 1}^{5} a_i \displaystyle \sum_{i = 2}^6 a_i$

## $\sum_{2}^{6} f(n)<\int_{1}^{6} f(x) d x<\sum_{1}^{5} f(n)$

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for the sake of demonstration we'LL just use this generic looking, continuous positive, decreasing function. The integral is thie entire area underneath the curve from one to six. In this case, we'Ll do that last this first Siri's, however, look something like this, rectangles of with one to make sure we have the whole valley of the height and the heights determined by the right side and points on the function. So for that first Siri's, which is from one to five, I mean over something looks roughly like this. For the last Siri's, however, we started to an end at six, and we can already see starting at two and ending at six. That the two Siri's are nearly identical, the primary difference being that the first Siri's has this large rectangle here in the last. Siri's instead has a much smaller rectangle over here s so we can conclude that the first serious is larger than the third. Siri's now when we need to do is think about how the integral compares Now the integral is only going from one, two six and again is going to be very much the same as the Siri's. In this case, it looks to be identical almost to the second Siri's. However, with this extra area up top, we can see that the area above the rectangles is not quite large enough to add up to the significant area of that first green rectangle, which includes area before these starting bound of are integral. So putting them in order from lowest toe largest, it would have the Siri's from two to six, then the Siri's from one Ah, my apologies, the integral from one to six and finally Siri's from one to five.

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